Problem: Let $h(x)=\begin{cases} \dfrac{3x}{xe^{x}}&\text{for }x\neq 0 \\\\ k&\text{for }x=0 \end{cases}$ $h$ is continuous for all real numbers. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $0$ (Choice C) C $e$ (Choice D) D $3$
Explanation: $\dfrac{3x}{xe^{x}}$ is continuous for all real numbers other than $x=0$ which means $h$ is continuous for all real numbers other than $x=0$. In order for $h$ to also be continuous at $x=0$, the following equality must hold: $\lim_{x\to 0}h(x)=h(0)$ Since $h(0)=k$, we will obtain the above equality by letting $k=\lim_{x\to 0}h(x)$. So let's find $\lim_{x\to 0}h(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 0}h(x) \\\\ &=\lim_{x\to 0}\dfrac{3x}{xe^{x}} \gray{\text{This is the rule for }x\neq 0} \\\\ &=\lim_{x\to 0}\dfrac{\cancel{3(x)}}{\cancel{(x)}(e^x)} \gray{\text{Factor}} \\\\ &=\lim_{x\to 0}\dfrac{3}{(e^x)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq 0) \\\\ &=\dfrac{3}{e^{0}} \gray{\text{Direct substitution}} \\\\ &=3 \end{aligned}$ We obtained that if we set $k=3$, then $\lim_{x\to 0}h(x)=h(0)$, which makes $h$ continuous at $x=0$. Since we already saw that $h$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $k=3$.